I am going through mathematical statistics with applications by Hogg, and he takes the partial derivates in respect to the parameters $\mu$ and $\sigma$.
For the following:
$$\ln L(\mu ,\sigma) = -\frac{n}{2}\ln (2\pi \sigma^2)-\frac{1}{2}\frac{1}{\sigma^2}\sum_{i=1}^n(y_i-\mu)^2$$
I am stuck on his solution of taking the partial derivative in respect to $\sigma^2$
He gets:
$$\frac{\partial \space \ln L(\mu, \sigma^2)}{\partial \space \sigma^2}=-\frac{n}{2}\frac{1}{\sigma^2}+\frac{1}{2}\left(\frac{1}{\sigma^2}\right)^2\sum_{i=1}^n(y_i-\mu)^2$$
I do not understand how he gets: $\frac{n}{2}\frac{1}{\sigma^2}$ - Where did the $\pi$ go? when I do it I get: $\frac{n}{2}\frac{1}{\pi \sigma^2}$
And then when he sets the derivative to 0, he finds that he gets:
$$-n\sigma^2+\sum_{i=1}^n(y_i-\mu)^2=0$$
However, when I simplify:
$$-\frac{n}{2\sigma^2}+\frac{1}{2\sigma^4}=\frac{-n\sigma^2+1}{2\sigma^2}$$
Unless I have gone wrong with my calculations?
First question - where does the $\pi$ go? (I ate it!) but joking aside $$ -\frac{n}{2}\ln(2\pi\sigma^2) = -\frac{n}{2}\ln(2\pi) -\frac{n}{2}\ln(\sigma^2) $$ The first term is independent of the variable that you are taking the derivative of so does not impact the location of the maxima/minima.
Not sure about your second question
$$-\frac{n}{2}\frac{1}{\sigma^2}+\frac{1}{2}\left(\frac{1}{\sigma^2}\right)^2\sum_{i=1}^n(y_i-\mu)^2 = \frac{1}{2\sigma^4} \left[-n\sigma^2 + \sum_{i=1}^n(y_i-\mu)^2\right] = 0$$
which is what the answer was.